Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{5 x + \left(x^{3} + \left(x^{2} - 5\right)\right)}{- 7 x^{2} + \left(\left(x^{2} + 1\right)^{2} - 3\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{5 x + \left(x^{3} + \left(x^{2} - 5\right)\right)}{- 7 x^{2} + \left(\left(x^{2} + 1\right)^{2} - 3\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{x^{3} + x^{2} + 5 x - 5}{x^{4} - 5 x^{2} - 2}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{- x^{3} - x^{2} - 5 x + 5}{- x^{4} + 5 x^{2} + 2}\right) = $$
$$\frac{-5 - 1^{2} - 1^{3} + 5}{- 1^{4} + 2 + 5 \cdot 1^{2}} = $$
= -1/3
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{5 x + \left(x^{3} + \left(x^{2} - 5\right)\right)}{- 7 x^{2} + \left(\left(x^{2} + 1\right)^{2} - 3\right)}\right) = - \frac{1}{3}$$